Sequences & Series
Harmonic Progression
Grade 11

Question:

<p><strong>(c)</strong> Let the positive numbers \(a, b, c, d\) be in A.P. Then \(abc, abd, acd, bcd\) are</p>
<p>(A) NOT in A.P./G.P./H.P.</p>
<p>(B) in A.P.</p>
<p>(C) in G.P.</p>
<p>(D) in H.P.</p>

Step-by-Step Solution

Key Concept: If four numbers are in A.P., their reciprocals form a sequence whose reciprocals can be analyzed. If terms are in H.P., their reciprocals are in A.P., so we need to check if the reciprocals of abc, abd, acd, bcd form an A.P.
<p><strong>Step 1:</strong> Given that a, b, c, d are in A.P., we can write them as:</p><p>a, a+r, a+2r, a+3r (where a > 0, r > 0 for positive terms)</p><p>So: a = a, b = a+r, c = a+2r, d = a+3r</p><p><strong>Step 2:</strong> The four terms given are abc, abd, acd, bcd. Let's find their reciprocals:</p><p>1/(abc) = 1/[a(a+r)(a+2r)]</p><p>1/(abd) = 1/[a(a+r)(a+3r)]</p><p>1/(acd) = 1/[a(a+2r)(a+3r)]</p><p>1/(bcd) = 1/[(a+r)(a+2r)(a+3r)]</p><p><strong>Step 3:</strong> To check if abc, abd, acd, bcd are in H.P., we verify if their reciprocals are in A.P.</p><p>Taking reciprocals:</p><p>1/(abc) = 1/[a(a+r)(a+2r)]</p><p>1/(abd) = 1/[a(a+r)(a+3r)]</p><p>1/(acd) = 1/[a(a+2r)(a+3r)]</p><p>1/(bcd) = 1/[(a+r)(a+2r)(a+3r)]</p><p><strong>Step 4:</strong> For A.P. verification, check: 1/(abd) - 1/(abc) = 1/(acd) - 1/(abd)</p><p>1/(abd) - 1/(abc) = 1/[a(a+r)] × [1/(a+3r) - 1/(a+2r)] = 1/[a(a+r)] × (-r)/[(a+2r)(a+3r)]</p><p>1/(acd) - 1/(abd) = 1/[a(a+3r)] × [1/(a+2r) - 1/(a+r)] = 1/[a(a+3r)] × (-r)/[(a+r)(a+2r)]</p><p><strong>Step 5:</strong> After simplification, these differences are equal, confirming the reciprocals form an A.P.</p><p><strong>∴ Answer: D</strong> The terms abc, abd, acd, bcd are in H.P.</p>
Correct Answer: D

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